Random "Dodecahedron" Walk

Relevant wiki: Linearity of Expectation

If we place the vertex at the origin, and the opposite vertex on the positive x-axis, then we can consider the following five groups of vertices (not including the initial vertex he's on) that share a common x-coordinate:

• Group 1: 3 points, all points neighboring the initial vertex
• Group 2: 6 points
• Group 3: 6 points
• Group 4: 3 points
• Group 5: the opposite vertex

Now if we consider the expectation values $E_n$ as the expected number of moves to get from any vertex in group $n$ to the vertex opposite the vertex he started on, then we get the following set of linear equations:

• $E=E_1 + 1$
• $E_1=1+(2/3)E_2+(1/3)E$
• $E_2 =1 + (1/3)E_1 + (1/3)E_2 + (1/3)E_3$
• $E_3=1+ (1/3)E_2+(1/3)E_3 + (1/3)E_4$
• $E_4=1+(2/3)E_3$

Solving for $E$ , you get $E=\boxed{35}$

Relevant wiki: Absorbing Markov Chains

I used the matrix method for Markov chains, equivalent to Geoff's solution: If

$R=\begin{bmatrix} 0&1/3&0&0&0\\1&0&1/3&0&0\\0&2/3&1/3&1/3&0\\0&0&1/3&1/3&2/3\\0&0&0&1/3&0\end{bmatrix}$

is the transition matrix between the states Geoff describes, including the initial position as State 0, but not including the "absorbing state" at the opposite vertex, then the answer is the sum of the entries in the first column of $(I_5-R)^{-1}=I_5+R+R^2+R^3+...$ , which comes out to be $\boxed{35}$ .